Question

In fig. 10.37, find tan P and cot R. Is tan P=cot R ?

Answer 1

using Pythagoras Law we got,

QR² = PR² - PQ²

$\Rightarrow QR=\sqrt{(}{PR}^2-{PQ}^2)\Rightarrow QP=\sqrt{(}{13}^2-{12}^2)\Rightarrow QP=\sqrt{(}169-144)\Rightarrow QP=\sqrt{25}\Rightarrow QP=5$

$\therefore tanP=\frac{QR}{PQ}$

$tanP=\frac{5}{12}and,cotR=\frac{QR}{PQ}cotR=\frac{5}{12}$

Now,

$tanP-cotR=\frac{5}{12}-\frac{5}{12}=0$